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Question Number 117811 by Lordose last updated on 13/Oct/20

$${\int }_0^{\mathit{\pi }}\mathbf{ln}\mid \mathbf{sinh}\left(\mathbf{x}\right)\mid \mathbf{dx}$$

Answered by mindispower last updated on 13/Oct/20

$$={\int }_0^{\pi }ln\left(sh\left(x\right)\right)dx$$$$sh\left(x\right)=\frac{e^x-e^{-x}}{2}$$$$={\int }_0^{\pi }(ln\left(e^x(1-e^{-2x})\right)-ln\left(2\right))dx$$$$={\int }_0^{\pi }(x+ln(1-e^{-2x})-ln\left(2\right))dx$$$$\text{Missing left or extra right}$$$$ln(1-e^{-2x})=-\sum _{n⩾0}\frac{e^{-2(n+1x}}{n+1}$$$${\int }_0^{\pi }ln(1-e^{-2x})dx=-\sum _{n⩾0}\frac{1}{n+1}{\int }_0^{\pi }{e}^{-2(n+)x}dx$$$$\sum _{n⩾0}\frac{e^{-2(n+1)\pi }-1}{2{(n+1)}^2}=\sum _{n⩾1}\frac{{\left(e^{-2\pi }\right)}^n}{n^2}-\sum _{n⩾1}\frac{1}{2n^2}$$$${Li}_s\left(x\right)=\sum _{k⩾1}\frac{x^k}{k^s}$$$$weget=\frac{1}{2}({Li}_2\left(e^{-2\pi }\right)-{Li}_2\left(1\right))=\frac{1}{2}{Li}_2\left(e^{-2\pi }\right)-\frac{1}{2}\zeta \left(2\right)$$$$=\frac{1}{2}{Li}_2\left(e^{-2\pi }\right)-\frac{{\pi }^2}{12}$$$$A=\frac{1}{2}{\pi }^2-\pi ln\left(2\right)$$$$A+B=\frac{5{\pi }^2}{12}-\pi ln\left(2\right)+{Li}_2\left(e^{-2\pi }\right)$$$${\int }_0^{\pi }ln\left(sh\left(x\right)\right)dx=\frac{5{\pi }^2}{12}-\pi ln\left(2\right)+{Li}_2\left(e^{-2\pi }\right)$$$$$$$$$$$$$$$$$$

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